Problem 38 - Chemistry

Question

(a) If a \(\mathrm{H}\) atom and a He atom are traveling at the same speed, what will be the relative wavelengths of the two atoms? (b) If a \(\mathrm{H}\) atom and a He atom have the same kinetic energy, what will be the relative wavelengths of the two atoms?

Step-by-Step Solution

Answer

In the case where Hydrogen and Helium atoms travel at the same speed, the wavelength of the hydrogen atom is four times that of the helium atom. However, if the Hydrogen and Helium atoms have the same kinetic energy, the wavelength of the Hydrogen atom is \(\sqrt{2}\) times the wavelength of the Helium atom.

Step 1: Applying de Broglie's equation

First, substitute the values of the Hydrogen atom and the Helium atom into the de Broglie wavelength formula \(\lambda = h/(mv)\). Let's assume that the speed is \(v\) for both.

Step 2: Determine the ratio of wavelengths

After calculations, \(\lambda_{H} = \frac{2h}{mv}\) for Hydrogen and \(\lambda_{He} = \frac{h}{2mv}\) for Helium. The ratio of these wavelengths is \(\lambda_{H} / \lambda_{He} = 4\). This means that the wavelength of the Hydrogen atom is four times the wavelength of the helium atom when they travel at the same speed.

Step 3: Determine the speed in terms of kinetic energy

For the second part, figure out the speed of each atom in terms of their kinetic energy. The kinetic energy \(E = 1/2 mv^2\), thus the speed \(v = \sqrt{2E/m}\). Replace \(v\) in the de Broglie equation for each atom.

Step 4: Compare the wavelengths

After substituting, \(\lambda_{H} = \frac{h}{\sqrt{2m_{H}E}}\) for Hydrogen and \(\lambda_{He} = \frac{h}{\sqrt{8m_{H}E}}\) for Helium (since \(m_{He} = 4m_{H}\)). The ratio between these wavelengths is \(\lambda_{H} / \lambda_{He} = \sqrt{2}\). Therefore, if they have the same kinetic energy, then the wavelength of the Hydrogen atom is \(\sqrt{2}\) times the wavelength of the Helium atom.

Key Concepts

Quantum Mechanics

Quantum Mechanics is a fundamental theory in physics that provides a description of the physical properties of nature at the scale of atoms and subatomic particles. It challenges classical notions by introducing probabilities rather than certainties into physical predictions.

One of the seminal discoveries of quantum mechanics is the de Broglie hypothesis that matter has wave-like properties. This concept is critical in understanding phenomena such as electron diffraction and the Heisenberg Uncertainty Principle. By relating the wavelength of particles to their momentum, de Broglie's equation bridges the gap between the microscopic quantum world and the macroscopic classical physics.

Atomic Physics

Atomic physics is the field of science that studies atoms as an isolated system of electrons and an atomic nucleus. This area of inquiry is concerned with processes such as ionization and excitation by photons or collisions with atomic and subatomic particles.

Within this arena, understanding the de Broglie wavelength of particles, including atoms like hydrogen and helium, provides insights into the structure of the atom and electron configurations. This knowledge is pivotal when discussing how atoms absorb and emit light, leading to the development of quantum models like the Bohr model and the quantum mechanical model of the atom.

Kinetic Energy

Kinetic energy is the energy that an object possesses due to its motion. It's a fundamental concept in physics, expressed in the equation \(E = \frac{1}{2} mv^2\), where \(E\) is the kinetic energy, \(m\) is the mass of the object, and \(v\) is its velocity.

In the context of quantum mechanics and atomic physics, kinetic energy is directly tied to the momentum of a particle, which in turn relates to the wavelength, as per de Broglie's formula. When comparing particles with the same kinetic energy but different masses such as hydrogen and helium atoms, it leads to an understanding that they will have different velocities and therefore, according to de Broglie's theory, different wavelengths.

Wavelength of Particles

The wavelength of particles is a quantum mechanical phenomenon, arising from the wave-particle duality which states that all particles exhibit both wave and particle properties. The de Broglie wavelength of a particle is given by \(\lambda = \frac{h}{p}\), where \(h\) is the Planck constant and \(p\) is the momentum of the particle.

This relationship demonstrates that the wavelength is inversely proportional to the particle's momentum. In our comparative exercise of hydrogen and helium atoms, it's shown that the lighter hydrogen atom will have a longer wavelength than helium when traveling at the same speed. Conversely, when they have the same kinetic energy, hydrogen's wavelength is still longer but by a different factor, due to the varying relationship between mass, kinetic energy, and velocity for each atom.